Problem: Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area. 

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

<image> Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold. 

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.

Examples

Input

3 2


Output

1.154700538379 1.632993161855


Input

2 100000


Output

70710.678118654752

Note

Definition of isosceles triangle: <https://en.wikipedia.org/wiki/Isosceles_triangle>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
template <class T>
T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
T lcm(T a, T b) {
  return (b / gcd(a, b)) * a;
}
void print() {}
void pLine() { cout << endl; }
template <typename T, typename... arg>
void print(T x, arg... ar) {
  cout << x;
  print(ar...);
}
const double PI = acos(-1);
const double eps = 1e-9;
using ld = long double;
void sol() {
  cout << fixed << setprecision(20);
  int n, h;
  cin >> n >> h;
  for (int i = (int)1; i < (int)n; i++) {
    ld res = sqrt((ld)i / (ld)n) * (ld)h;
    print(res, ' ');
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  sol();
  return 0;
}